Question

(a) Knowing that the moment of inertia of a thin uniform metallic rod of mass m and length L about an axis through its center of mass is (1/12) mL?. what is its moment of inertial about a parallel axis through one of its ends (show your calculation). (b) A physical pendulum consisting of a thin metallic rod of mass m = 200.0 g and of length L - 1.000 m is suspended from the upper end by a frictionless pivot. It makes 50 complete swings in 87.04 s. What is the acceleration of gravity at the location of the pendulum? (e) At what altitude above the Earth's surface is this pendulum placed? (18 points) TT T Arial T.EE 3 (12pt)

Answer 1

I le = ML ² Parallel axis theorm using M T- Ict MLB ML? -> + 12 L 3 13 Time Physizat pendulum Perred of Tor ma 1 - moment of inertia about plvot de distance of pivot from centre ។ nes M22 I = m² 87.04 50 3. mg. 12 3 No. of cheles time 1.74 - 2018 22 3 (107472 1 472 X 2 3 (1.9472 g'= 8. 69 m/s2 rea --> GM RL Re 9L => 1.06R h = 9.8. R Height from he coby 6.4x106 earth surface (2) h = 1.06R-R - = 0.384x 10 m & R 0.06R